find $\\frac{dy}{dt}$. $y = 3t(2t^{2}-5)^{4}$ $\\frac{dy}{dt}=\\square$
find $\\frac{dy}{dt}$. $y = 3t(2t^{2}-5)^{4}$ $\\frac{dy}{dt}=\\square$
Answer
Explanation:
Step1: Apply the product rule
The product rule states that if (y = uv), then (\frac{dy}{dt}=u\frac{dv}{dt}+v\frac{du}{dt}). Let (u = 3t) and (v=(2t^{2}-5)^{4}). First, find (\frac{du}{dt}): (\frac{du}{dt}=\frac{d}{dt}(3t)=3) Next, find (\frac{dv}{dt}) using the chain rule. Let (w = 2t^{2}-5), so (v = w^{4}). The chain rule: (\frac{dv}{dt}=\frac{dv}{dw}\cdot\frac{dw}{dt}) (\frac{dv}{dw}=4w^{3}) and (\frac{dw}{dt}=4t) So (\frac{dv}{dt}=4(2t^{2}-5)^{3}\cdot4t = 16t(2t^{2}-5)^{3})
Step2: Substitute into the product rule formula
(\frac{dy}{dt}=u\frac{dv}{dt}+v\frac{du}{dt}) (\frac{dy}{dt}=3t\cdot16t(2t^{2}-5)^{3}+(2t^{2}-5)^{4}\cdot3) (=48t^{2}(2t^{2}-5)^{3}+3(2t^{2}-5)^{4}) Factor out (3(2t^{2}-5)^{3}): (=3(2t^{2}-5)^{3}[16t^{2}+(2t^{2}-5)]) (=3(2t^{2}-5)^{3}(18t^{2}-5))
Answer:
(3(2t^{2}-5)^{3}(18t^{2}-5))